Transcript Document

Quiz
Determine the minimum number of shift register stages required to create a
maximal length PN sequence which has a repetition time greater than 10 seconds
and a chip rate of 1.23 Mc/s.
12.3 x 106  2 N  1
N  log 2 (12.3 x 106  1)  23.5  24 stages
Multiple Access Systems
FDMA
30 kHz FM channels – 900 Mhz band
#users NT = Bsys/Bch
TDMA
3:1 compression of Old FDMA channels
CDMA
2/4 PSK
1.23 Mhz Spreading codes
Transmit Power control to equalize incoming power levels
Cellular Protocol
Number of cells in cluster
Reuse factor Kr = a2 + ab + b2
dI
Kr = 173
Ncell 
r
3
Bsys
2
3
Bch Kr
2
1
5
7
d 
0.3  I 
 r 
5
7
6
3
6
Kr
1
4
Cell Radius r
For Co-channel
Separation distance dI
4
2
2
C
d 
 0.17  I 
I0
 r 
2
4
4
1
Cellular CDMA Capacity
Since all cells use the same bandwidth, and mobile unit power is
controlled, it is only the presence of other users that limit capacity.
NCDMA
 G  1   


 
E
/
I
1


 v 
 b 0 
For Eb/I0 = 7 dB, NCDMA = 25
G = Bs/Rv
Bs = Spreading Chip Rate ~ 1.23 Mb/s
Rv = Voice Bit Rate ~ 9.6 kb/s
 = 0.5 Neighboring cell interference
 = 0.85 Power Control Factor
v = 0.6 Voice Activity factor
CDMA is much less susceptible to fading than narrowband FM
because of broad bandwidth used.
Cellular Radio Propagation
Path Loss
Doppler (for texting while driving)
Slow log-normal fading
Fast multipath fading
30-40 dB variations over fractional wavelengths
Time dispersion
Burst errors
Rayleigh statistics if LOS is not dominant
1  2
Pe  e
2
 =C N
Rayleigh multipath + Dominant LOS > Rician fading statistics
Pe 
1


2



k

Note that Rice
e
 
  2 
 k
Rayleigh as k
k=
0
LOS
MP
Digital Transmission
n bits/symbol
M = 2n channel states (phasors)
Channel must switch between states every Ts = 1/fs seconds,
Thus fi = nfs
Bandwidth efficiency
fi
B
unitless (Bits/sec per hz)
Since there are n bits per symbol, Es = n Eb and fs Es = fi Eb
For M-PSK, phase noise required for an error is
1  2  
 N  

2 M  M
And the noise voltage to signal voltage for an error would be
Vn Vˆs sin   N 

 sin   N 
Vˆs
Vˆs
Probability of Error
Vˆs  2Vs , RMS is the magnitude of the symbol phasor.
VˆS
VN , RMS

2VS , RMS
VN , RMS

2
S , RMS
2
N , RMS
2V
V
 2 Es 


N
 0 
1
2
The probability of symbol error is given by
 2 E  12



s
PE ( M )  2Erfc 
 sin    
N0 





Erfc(x) is the complimentary error
function for Gaussian Distributions:
The probability of bit error is
PE ( M )
Pb 
n
Erfc( x)  

x
e
u 2
2
 du
e
 x2
2

x 2
Probability of Error (cont)
For M-FSK,
M
B  Mf s 
Ts
The probability of symbol error is
M  1   Es 2 N0  2n  1   Es 2 N0 
PE  M  
e

e
2
2
PE  M 
Pb 
n
Viterbi sez . . .
2n 1
Pb  n
PE
2 1